Find if a number contains a digit without ever using using its string representation

Question

The Challenge:

Find if a number contains a digit. The number and the digit are to be taken as input. Output must be a boolean value. But you are not allowed to use the string representation of the number at any time. Now don't count upon a char* either. You must devise a mathematical way to do it.

Rules:

• You can't use a string/char[] representation of the number. Same goes for the digit.

• It must be a complete program.

• Input can be in any form.

• Output must be a boolean value.

This is code-golf, so shortest code wins!.

Test Cases:

Input: 51, 5
Output : True
------------------------
Input: 123456789, 0
Output : False
------------------------
Input: 1111, 1
Output : True

Disclaimer: I don't understand much Pyth/CJam/Japt and similar languages. Thus I hope everyone will be honest enough not to break the rules. Never mind, the community is always there to help me :)

Answer 1

Perl 6, 27 bytes

{$^a.polymod(10 xx*)∋$^b}

Usage:

# make it an infix op, because I can
my &infix:<contains> = {$^a.polymod(10 xx*)∋$^b}

say 51 contains 5; # True
say 123456789 contains 0; # False
say 1111 contains 1; # True

# you can use infix ops as oddly named subs as well
say infix:<contains> 1230, 0; # True

say map {$^a.polymod(10 xx*)∋$^b}, 10,0, 12,3, 12,2;
# (True False True)

# proof it isn't doing anything string related
'hi' contains 'h';
# Method 'polymod' not found for invocant of class 'Str'

# Since it uses a Set operator the second argument has to be an Int
# not a Str
say 123 contains '1'; # False
# not a Rat
say 123 contains 1.0; # False
# not a Num
say 123 contains 1e0; # False
# but an Int
say 123 contains 1; # True

Explanation:

polymod is similar to divmod from other languages, except it takes the number and a list of divisors. Here I give it an endless list of 10s so that it splits it up into a list of digits.

Then I see if the resulting list contains the second argument by using the 'contains as member' ( ∋ / (cont) ) Set operator.

Answer 2

Haskell, 42 bytes

Called like number!digitToFind:

n!d|n<d=1>2|mod n 10==d=1<2|1<2=div n 10!d

Answer 3

MATL, 15 14 bytes

t`10H#\t]xN$hm

Try it online!

t                   % take 1st input (number) implicitly. Duplicate
 `      ]           % do...while loop
  10                % push number 10, to be used as divisor
    H#\             % divmod. Pushes remainder and then quotient
       t            % duplicate quotient. Used as loop condition: exit if zero
         x          % delete last quotient (which is zero)
          N$h       % concatenate all partial results in an array
             =      % take 2nd input (digit) implicitly.
              m     % ismember function. True if digit is in array

Answer 4

Python 3, 59

It uses the same approach as every other approach.

def f(x,s):
 while x:
  x,r=divmod(x,10)
  if r==s:return 1

Answer 5

Befunge-93, 38 bytes

&&00pv >[email protected]<
%*25:<_^#:/_ ^#\*25-g00

Arguments are taken in the form MainNum Digit. It's the same algorithm that everyone else is doing (check if mod 10 is equal to the digit; if not, div 10 until it equals 0), since that's pretty much the optimal way to do this one.

Answer 6

JavaScript (ES6), 30 bytes

d=>g=n=>n%10==d||n>9&&g(n/10|0)

Examples:

f=d=>g=n=>n%10==d||n>9&&g(n/10|0)
f(0)(0) -> true
f(0)(10) -> true
f(0)(12) -> false
f(1)(12) -> true
f(2)(12) -> true
f(3)(12) -> false

Answer 7

PHP, 66 65 62 47 bytes

in_array($d,array_map('intval',str_split($n)));

$n is the number to be searched
$d is the digit to search for

This code converts the number into array[number], then scans the array for a matching digit. Returns true if any digit matches.

Alternatively, if number->string->array[string]->array[number] is not valid (even though no operations use it as a string), there's this at 62 bytes, for any $n > 0:

function f($n,$d){return$n&&($d==$n%10||f(($n-$n%10)/10,$d));}